Talk:Inaccessible cardinal
Fixed points What does \psi_I(1) mean? Does it add another \Omega at the bottom? (As in \underbrace{\Omega_{\Omega_{\Omega_{\Omega_{...\Omega}}}}}_{\omega+1} ) King2218 (talk) 15:33, March 11, 2014 (UTC) No. \(\psi_I(1) = lim\{\psi_I(0), \Omega_{\psi_I(0)+1}, \Omega_{\Omega_{\psi_I(0)+1}}, ...\}\) Wythagoras (talk) 15:30, March 11, 2014 (UTC) Argh, why do I always forget how to escape the fixed point trap?! Anyways, thanks! King2218 (talk) 15:33, March 11, 2014 (UTC) More questions :P 1. How does one access the inaccessible cardinal? (or, at least, how huge is it?) 2. \Omega is the diagonalizer of the \psi function which makes large ordinals and I is the diagonalizer of the \psi_I which makes large uncountable ordinals. What are the next terms of the sequence? King2218 (talk) 06:54, March 19, 2014 (UTC) :1. It is larger than anything "constructible" in ZFC. In particular, it is larger than anything you can construct using the Aleph or Beth functions, extended up using arbitrary hierarchies. So, just as we can be sure that when we collapse anything using \psi_\Omega for any of our extensions (now or in the future), we get an ordinal less than \Omega , we can be sure that any of our extensions of \psi_I (now or in the future) will be less than the smallest inaccessible (in fact, smaller than the smallest weakly inaccessible) :2. The sequence goes \Omega, I, I_2, I_3, \ldots . We diagonalize over that sequence by using the smallest Mahlo cardinal M'' and the \chi function: e.g. \chi(\alpha) = I_\alpha, \chi(M) diagonalizes over the function \alpha \mapsto I_\alpha , and so on. Always omega fixed? Is any inaccessible cardinal necessarily omega fixed? -- 22:15, February 26, 2015 (UTC) If you mean is every inaccessible cardinal a fixed point of the function \(\alpha \mapsto \Omega_\alpha\), the answer is yes, even for weakly inaccessible cardinals. This follows from the fact that a weakly inaccessible cardinal is a regular and limit cardinal. If \(I\) is weakly inaccessible, and \(I = \Omega_\alpha\), then \(\alpha\) must be a limit ordinal, since \(I\) is a limit cardinal. If \(\alpha < I\), then the set \(\lbrace \Omega_\beta | \beta < \alpha \rbrace\) would be a an unbounded subset of \(I\) of size \(\alpha\), which would mean the cofinality of \(I\) is at most \(\alpha\), which would contradict the regularity of \(I\). (An ordinal \(\beta\) is regular if the cofinality of \(\beta\) is \(\beta\).) Of course we cannot have \(\alpha > I\), as the function \(\Omega_\alpha\) is strictly increasing. So \(I = \Omega_I\). Note that the converse is not true - the smallest cardinal \(\alpha\) such that \(\alpha = \Omega_\alpha\) is not weakly inaccessible, as it is the limit of \(\Omega, \Omega_\Omega, \Omega_{\Omega_\Omega}, \ldots\), and therefore has cofinality \(\omega\). In fact the smallest weakly inaccessible cardinal lies far beyond anything we can construct using \(\Omega\) fixed points and any further diagonalizations of that, which makes weakly inaccessibles ideal for collapsing down to \(\Omega\) fixed points. Deedlit11 (talk) 02:09, February 27, 2015 (UTC) Another question Is any strongly inaccessible cardinal also weakly inaccessible? -- 09:28, February 27, 2015 (UTC) : Yes - every strong limit cardinal is also a normal limit, so every strongly inaccessible is also weakly inaccessible. LittlePeng9 (talk) 10:10, February 27, 2015 (UTC) Continuum Is it "possible" that the first '''weakly' inaccessible cardinal = the continuum? (Since the continuum hypothesis is independent with ZFC) If yes, then is it "possible" that the first weakly Mahlo cardinal = the continuum? {hyp/^,cos} (talk) 02:38, July 27, 2017 (UTC) :I believe the answers are yes and yes. Solovay's theorem tells us that, if \(\alpha\) is any cardinal with cofinality > \(\omega\), then it is consistent that \(2^{\aleph_0} = \alpha\). More precisely, if we start with a model M of ZFC, one can find a generic extension of N of M such that in N, \(2^{\aleph_0} = \alpha\). So we can answer your first question in the affirmative by picking \(\alpha\) to be the first weakly inaccessible cardinal - assuming that the property of weak inaccessibility is invariant under forcing. Since in Solovay's and Easton's theorems, cofinality is referenced prior to the model being chosen, it would appear that cofinality, and therefore regularity, is model independent; so the first weakly inaccessible cardinal will remain the first weakly inaccessible cardinal after forcing, and the answer to your first question is yes. :For the second question, the situation is the same - it comes down to whether the property of being weakly Mahlo is independent of forcing. For this I don't have the previous argument, but I suspect the answer to this question is also yes. Deedlit11 (talk) 03:10, July 27, 2017 (UTC) ::So it's possible that the first strongly compact cardinal = the continuum. 16:26, December 27, 2017 (UTC) :::No. s are measurable, hence they are strong limits. Continuum is not a strong limit. LittlePeng9 (talk) 16:37, December 27, 2017 (UTC) ::::This means the continuum